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Thread: sequence

  1. #1
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    sequence

    Prove the following: Let $\displaystyle (a_n) $ and $\displaystyle (b_n) $ be sequences of real numbers that converge to real numbers $\displaystyle a $ and $\displaystyle b $ respectively. If $\displaystyle a_n \leq b_n $ for all $\displaystyle n \in \mathbb{N} $, then $\displaystyle a \leq b $.

    Proof: For each positive real number $\displaystyle \epsilon $ there exists a $\displaystyle N_1 \in \mathbb{N} $ so that whenever $\displaystyle n > N_1 $, $\displaystyle |a_n-a| < \epsilon $. Also for each positive real number $\displaystyle \epsilon $ there exists a $\displaystyle N_2 \in \mathbb{N} $ so that whenever $\displaystyle n > N_2 $, $\displaystyle |b_n-b| < \epsilon $. So $\displaystyle a- \epsilon < a < a + \epsilon $ and $\displaystyle b- \epsilon < b < b+ \epsilon $. Then $\displaystyle 0 \leq b_n- a_n < b-a $ for all $\displaystyle n \in \mathbb{N} $. It follows that $\displaystyle a \leq b $. $\displaystyle \square $

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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Prove the following: Let $\displaystyle (a_n) $ and $\displaystyle (b_n) $ be sequences of real numbers that converge to real numbers $\displaystyle a $ and $\displaystyle b $ respectively. If $\displaystyle a_n \leq b_n $ for all $\displaystyle n \in \mathbb{N} $, then $\displaystyle a \leq b $.

    Proof: For each positive real number $\displaystyle \epsilon $ there exists a $\displaystyle N_1 \in \mathbb{N} $ so that whenever $\displaystyle n > N_1 $, $\displaystyle |a_n-a| < \epsilon $. Also for each positive real number $\displaystyle \epsilon $ there exists a $\displaystyle N_2 \in \mathbb{N} $ so that whenever $\displaystyle n > N_2 $, $\displaystyle |b_n-b| < \epsilon $. So $\displaystyle a- \epsilon < a < a + \epsilon $ and $\displaystyle b- \epsilon < b < b+ \epsilon $. Then $\displaystyle 0 \leq b_n- a_n < b-a $ for all $\displaystyle n \in \mathbb{N} $. It follows that $\displaystyle a \leq b $. $\displaystyle \square $

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    Try proving the following easier result: let $\displaystyle \{ x_n\}$ be a sequence such that $\displaystyle x_n \geq 0$ and $\displaystyle x_n \to x$ then $\displaystyle x\geq 0$.

    After you prove that consider $\displaystyle x_n = a_n - b_n \geq 0$.
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